Question

which expression is a cube root of 2i

a. cube root of 2 (cos (210°) + i sin (210°))
b. cube root of 2 (cos (260°) + i sin (260°))
c. cube root of 2 (cos (90°) + i sin (90°))
d. cube root of 2 (cos (60°) + i sin (60°))

Answer 1

Do you know demoivres?

Answer 2

No I dont

Answer 3

Ok so cos (theta) + isin(theta)=\[e ^{thetai}\]

Answer 4

so first we express i in terms of cos(theta) +isin(theta

Answer 5

how? like 2cos(theta) +2 i sin (theta) ?

Answer 6

Ok I got D

Answer 7

Ill draw it

Answer 8

D is not correct, but it's close

Answer 9

oh wait, lol wrong mode, my bad

Answer 10

lol nice 99 dude ^

Answer 11

one sec, I gotta redo

Answer 12

so its D?

Answer 13

Yea do you want me to teach you Demoivres ?

Answer 14

yeah please. How did you get that?

Answer 15

Awesome so i on the imaginary axis is (1, pi/2)

Answer 16

so we can write i in terms of polar coordinates

Answer 17

as cos (pi/2) +isin(pi/2)

Answer 18

oh okay I think I got it now . Thanks

Answer 19

The problem is this

if z = cube root of 2 (cos (60°) + i sin (60°))

then


z = cube root of 2 (cos (60°) + i sin (60°))

z^3 = (cube root of 2)^3 * ( cos(60*3) + i*sin(60*3) )

z^3 =2 * ( cos(180) + i*sin(180) )


but that will land you on the complex number 2+0i which is NOT 0+2i

Answer 20

so this shows us that cube root of 2 (cos (60°) + i sin (60°)) is not a cube root of

2 (cos (90°) + i sin (90°))

Answer 21

oops I meant to say -2+0i, but still, that's not 0+2i

Answer 22

??? why did you multiply by 3?

Answer 23

because

if

z = r( cos(theta) + i*sin(theta) ),

then

z^n = r^n*( cos(n*theta) + i*sin(n*theta) )


by De Moivre's theorem

Answer 24

in this case, n = 3 because we're looking at cube roots

Answer 25

ummm no

Answer 26

yes, z is a cube root of z^3

Answer 27

I'm checking answer choice D by cubing it

You should be able to cube choice D and get 2 (cos (90°) + i sin (90°))